Thanks for the video, but I have some questions: the Hessian matrix seems to be a function of x and so we still have not resolved the problem of how to evaluate it across the whole domain? How do you evaluate the principal minors without knowing at what point x to evaluate them? And if you cannot evaluate the principal minors we cannot apply the rules learnt to characterize the Hessian matrix. Please how do we use the Hessian matrix, hopefully with a concrete example, to show that the function is concave? Thanks and hope to hear from you soon
Yes, the Hessian matrix certainly depends on x. For a specific function we obtain expressions for the first and second derivatives (if possible) and we try to determine whether the second-derivative principal-minor expressions have the right signs over the whole domain. For example, for the real function f(x) = 60x - 3x^2, we have f"(x) = -6 everywhere, so f is strictly concave and is maximized at x = 10, where f'(x) = 0. For the function f(x_1,x_2) = 80 - x_2^2 - x_1^2 + 8x_1 on R^2, we have a Hessian matrix H in which f_11 = f_22 = -2 and f_12 = f_21 = 0, so H is negative definite everywhere, and therefore strictly concave; f is maximized at x = (4,0), where the gradient is the zero vector. For an example in which the second derivative is not constant, try the real function f(x) = 50 log x - x^2 on the positive reals; you'll get f"(x) < 0 everywhere and x is maximized at x = 5.
@@ArizonaMathCamp Thanks for the detailed quick response. Much appreciated. I understand the examples you provided, but they did not solve the issue raised about the Hessian matrix depending on x. For example, if I get a Hessian with f_11 = 2x, f_22 = -3y, and f_12 = f_21 = 2xy, I will usually evaluate the Hessian at the critical points to understand whether the critical points are a local maximum or minimum. But how would I evaluate this Hessian across the whole R^2 domain to check whether the original function is concave or convex and hence critical points can be taken as either global maximum or minimum? My difficulty is in using the theorems to identify a concave or convex function from the Hessian and hence establish whether the critical points are global points and not just local points. Hope to hear from you soonest. Thanks a million for your generosity
@@ArizonaMathCamp For the function f(x) = 50 log x - x^2, the f''(x) = - 50/x^2 - 2 and hence the square ensures that the value and sign of x does not change the sign of the sum, which is still negative. Hence the original function is concave and the critical point x = 5 gives a global maximum. For a Hessian matrix with terms that are functions of several variables, it seems difficult to evaluate the Hessian across the whole domain. I need a way out to deal with these kinds of complex problems. Thanks
@@tayoukachukwu3105 In your example det H < 0 on all of the positive quadrant of R^2 so it's neither concave nor convex there. Same on the negative quadrant. This generally means that the local extrema are not global extrema.
@@ArizonaMathCamp Thanks for the response. Much appreciated. In essence you are suggesting that I have no option but to evaluate the Hessian choosing examples from four quadrants of the domain; if the different options satisfy the conditions for negative semi-definite or positive semi-definite then I can conclude concave or convex function respectively and then critical points can be viewed as global extrema. I hope I understood you correctly. Thanks
U of A students are very lucky!
Thanks for the video, but I have some questions: the Hessian matrix seems to be a function of x and so we still have not resolved the problem of how to evaluate it across the whole domain? How do you evaluate the principal minors without knowing at what point x to evaluate them? And if you cannot evaluate the principal minors we cannot apply the rules learnt to characterize the Hessian matrix. Please how do we use the Hessian matrix, hopefully with a concrete example, to show that the function is concave? Thanks and hope to hear from you soon
Yes, the Hessian matrix certainly depends on x. For a specific function we obtain expressions for the first and second derivatives (if possible) and we try to determine whether the second-derivative principal-minor expressions have the right signs over the whole domain. For example, for the real function f(x) = 60x - 3x^2, we have f"(x) = -6 everywhere, so f is strictly concave and is maximized at x = 10, where f'(x) = 0. For the function f(x_1,x_2) = 80 - x_2^2 - x_1^2 + 8x_1 on R^2, we have a Hessian matrix H in which f_11 = f_22 = -2 and f_12 = f_21 = 0, so H is negative definite everywhere, and therefore strictly concave; f is maximized at x = (4,0), where the gradient is the zero vector. For an example in which the second derivative is not constant, try the real function f(x) = 50 log x - x^2 on the positive reals; you'll get f"(x) < 0 everywhere and x is maximized at x = 5.
@@ArizonaMathCamp Thanks for the detailed quick response. Much appreciated. I understand the examples you provided, but they did not solve the issue raised about the Hessian matrix depending on x. For example, if I get a Hessian with f_11 = 2x, f_22 = -3y, and f_12 = f_21 = 2xy, I will usually evaluate the Hessian at the critical points to understand whether the critical points are a local maximum or minimum. But how would I evaluate this Hessian across the whole R^2 domain to check whether the original function is concave or convex and hence critical points can be taken as either global maximum or minimum? My difficulty is in using the theorems to identify a concave or convex function from the Hessian and hence establish whether the critical points are global points and not just local points. Hope to hear from you soonest. Thanks a million for your generosity
@@ArizonaMathCamp For the function f(x) = 50 log x - x^2, the f''(x) = - 50/x^2 - 2 and hence the square ensures that the value and sign of x does not change the sign of the sum, which is still negative. Hence the original function is concave and the critical point x = 5 gives a global maximum. For a Hessian matrix with terms that are functions of several variables, it seems difficult to evaluate the Hessian across the whole domain. I need a way out to deal with these kinds of complex problems. Thanks
@@tayoukachukwu3105 In your example det H < 0 on all of the positive quadrant of R^2 so it's neither concave nor convex there. Same on the negative quadrant. This generally means that the local extrema are not global extrema.
@@ArizonaMathCamp Thanks for the response. Much appreciated. In essence you are suggesting that I have no option but to evaluate the Hessian choosing examples from four quadrants of the domain; if the different options satisfy the conditions for negative semi-definite or positive semi-definite then I can conclude concave or convex function respectively and then critical points can be viewed as global extrema. I hope I understood you correctly. Thanks